Question

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Answer 1

Answer:

$x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}$

Step-by-step explanation:

Biquadratic Equations

Solve:

$x^4 + 3x^2 - 18 = 0$

The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.

Solving such equations requires to express the equation as a second-degree equation with $x^2$ as the variable.

Rewriting the equation:

$(x^2)^2 + 3(x^2) - 18 = 0$

The quadratic equation can be factored as:

$(x^2+6)(x^2-3)=0$

It leads to two equations:

$x^2+6=0$

$x^2-3=0$

The first equation has imaginary roots. Solving for x:

$x^2=-6$

$x=\pm\sqrt{-6}$

$x_1=\mathbf{i}\sqrt{6}$

$x_2=-\mathbf{i}\sqrt{6}$

Where

$\mathbf{i}=\sqrt{-1}$

The second equation has two real roots:

$x^2-3=0$

$x^2=3$

$x=\pm\sqrt{3}$

$x_3=\sqrt{3}$

$x_4=-\sqrt{3}$

The roots are:

$\mathbf{x_1=\mathbf{i}\sqrt{6},\ x_2=-\mathbf{i}\sqrt{6},\ x_3=\sqrt{3},\ x_4=-\sqrt{3}}$